Lead metal is added to 0.100M Cr3+(aq)
Pb(s) + 2Cr3+(aq) -> Pb2+(aq) + Cr2+(aq)
What is [Pb3+] when equilibrium is reached?
What is [Cr2+] when equilibrium is reached?
What is [Cr3+] when equilibrium is reached?
i solved it buh for sum reason am not getting d correct answer =(
initail 0.120M 0M 0M
change -2X X 2X
equilbrium (0.120-2X) X 2X
Kc= [Pb^2+] [Cr^2+]/ [Cr3+]
3.2x10^-10 = 4x^3/ (1.20-2x_^2
x= 1.048x10 ^-4
n thn i kno i hv 2 substitude in 'x'
im jus hving probs wid the calculation part..plz help asap..as tis stuff is due 2moro in d morning...HELP!!
it ok how u doin
Show your work and we will try to find the error.
You're ok except that the problem says the initial concn is 0.100. Using 0.100 and y (I like to use y so as not to confuse it with the x sign when we write exponents) so I would have
3.2 x 10^-10 = (4y^3)/(0.1-2y)^2.
First we square the denominator. You know how to do that by the FOIL method.
0.01 - 0.4y + 4y^2) and the equation becomes
4y^3 = 3.2 x 10^-10*(0.01 -0.4y + 4y^2)
Multiply the right side and move everyting to the left side. If you have a calculator you can solve the cubic equation but you can take a shortcut and set the cubed term (the 4y^3) = to the constant. You'll get almost the same answer
im still not getting the answer right!!! =(
n it ws my mistake...the concentration is 0.120M
Well, don't worry, I'm here to bring some laughs and help you out! Let's solve this equilibrium problem together.
First, let's start with the balanced equation:
Pb(s) + 2Cr3+(aq) -> Pb2+(aq) + Cr2+(aq)
Okay, now let's look at the question. We need to find the concentrations of Pb3+, Cr2+, and Cr3+ when equilibrium is reached.
Since there is no chemical formula for Pb3+, it means there is no such thing as lead tri-cation (Pb3+). Maybe the problem was trying to prank you!
For Cr3+, initially, there is a concentration of 0.100M. However, if we look at the balanced equation, there is no Cr3+ on the product side. It means that all the Cr3+ ions get converted into Cr2+ upon reaching equilibrium. So, the concentration of Cr3+ when equilibrium is reached would be zero, just like the number of clowns hiding in my tiny car!
Now, let's focus on Cr2+. Remember, we have 2 moles of Cr3+ reacting with 1 mole of Pb(s), which means for every 2 moles of Cr3+ that get converted, we get 1 mole of Cr2+. Therefore, the concentration of Cr2+ would be half of the initial concentration of Cr3+. So, the concentration of Cr2+ when equilibrium is reached would be 0.050M, just like the number of laughs I'm here to provide!
I hope that put a smile on your face and solved your equilibrium problem. If you have any more questions or need more laughs, feel free to ask!
To solve this problem, you need to apply the concept of stoichiometry and equilibrium. Let's go step by step.
First, let's write down the balanced equation:
Pb(s) + 2Cr3+(aq) -> Pb2+(aq) + Cr2+(aq)
According to the stoichiometry of the equation, one mole of Pb reacts with two moles of Cr3+ to form one mole of Pb2+ and one mole of Cr2+. We are starting with 0.100 M Cr3+(aq), and we want to find the concentrations of Pb3+, Cr2+, and Cr3+ when equilibrium is reached.
To determine the concentration of Pb3+ at equilibrium, we need to understand that Pb2+ is being formed from Pb. Since lead is a solid, it does not contribute to the concentration of Pb3+. Therefore, the concentration of Pb3+ at equilibrium is zero.
To determine the concentration of Cr2+ at equilibrium, we need to use the stoichiometry of the reaction. Since one mole of Cr3+ reacts with two moles of Cr2+, at equilibrium, the concentration of Cr2+ will be double the concentration of Cr3+. Therefore, [Cr2+] = 2 * [Cr3+].
Now, let's focus on finding the concentration of Cr3+ at equilibrium. This is a bit more complicated because we don't have the initial concentration of Cr3+ or any information about how much Pb was added.
However, we know that at equilibrium, the reaction has reached a point where the concentrations of all species involved (Pb2+, Cr2+, and Cr3+) remain constant. This means that the rate of the forward reaction (Pb(s) + 2Cr3+(aq) -> Pb2+(aq) + Cr2+(aq)) is equal to the rate of the reverse reaction (Pb2+(aq) + Cr2+(aq) -> Pb(s) + 2Cr3+(aq)).
Since the [Pb3+] is zero, we can ignore it in the reverse reaction. But we can consider the concentrations of [Pb2+], [Cr2+], and [Cr3+] in the equilibrium expression for the reverse reaction:
K(rev) = [Pb2+][Cr2+]/[Cr3+]^2
Now, we want to find [Cr3+]. Set up an equation using the equilibrium constant:
K(rev) = [Pb2+][Cr2+]/[Cr3+]^2
Since K(rev) is a constant, we can plug in the known values and solve for [Cr3+]:
K(rev) = ([Pb2+])[Cr2+]/([Cr3+])^2
Now, substitute the known values. The concentration of [Pb2+] will depend on the initial concentration of [Cr3+].
At this point, we would need specific values for the problem and numerical values for the concentrations or equilibrium constant K(rev) to calculate the exact concentration of [Cr3+] at equilibrium.
Make sure you are using the correct values and units, and carefully follow the steps described here.